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JEE Mains · Maths · STD 12 - 9. differential equations

વિધેય \(\mathrm{F}:[3,5] \rightarrow \mathrm{R}\) એ અંતરાલ \((3,5)\) પર બે વખત વિકલનીય છે કે જેથી  \(\mathrm{F}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}}\) \(\int_{3}^{x}\left(3 t^{2}+2 t+4 F^{\prime}(t)\right) \,d t\) થાય. જો \(F^{\prime}(4)=\frac{\alpha e^{\beta}-224}{\left(e^{\beta}-4\right)^{2}}\) તો  \(\alpha+\beta\) ની કિમંત મેળવો.

  1. A \(8\)
  2. B \(16\)
  3. C \(48\)
  4. D \(32\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(16\)

Step-by-step Solution

Detailed explanation

\(F(3)=0\) \(\mathrm{e}^{\mathrm{x}} \mathrm{F}(\mathrm{x})=\int_{3}^{\mathrm{x}}\left(3 \mathrm{t}^{2}+2 \mathrm{t}+4 \mathrm{~F}^{\prime}(\mathrm{t})\right) \,\mathrm{dt}\) \(e^{x} F(x)=e^{x} F^{\prime}(x)=3 x^{2}+2 x+4 F^{\prime}(x)\)…
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